IB Mathematics AI AHL Vector applications to kinematics MAI Study Notes- New Syllabus

IB Mathematics AI AHL Vector applications to kinematics MAI Study Notes

LEARNING OBJECTIVE

Vector applications to kinematics.

Key Concepts:

Kinematics with Vectors
Constant & Variable Velocity

A nice application of the vector equation of a line is the following:

◆ VELOCITY AND SPEED

Suppose that a body is moving along a straight line with a constant velocity, and its position at time $ t $ is given by:

$\vec{r} = \vec{a} + t\vec{b}$

Then:

$\vec{a}$ is the position of the body at time $ t = 0 $.
$\vec{b}$ is the velocity vector of the body (usually denoted as $\vec{v}$).
$|\vec{b}|$ is the speed of the body (usually denoted as $|\vec{v}|$).

The vectors (and thus the motion) can be in either 2D or 3D space.

Example

Suppose that a body is moving according to the equation:

$\vec{r} = \begin{pmatrix} 1 \\ 2 \end{pmatrix} + t \begin{pmatrix} 3 \\ 4 \end{pmatrix}$
where time is measured in seconds and distance in meters.

The initial position (at $ t = 0 $)
The position at time $ t = 1 \, \text{sec} $
The velocity vector is $\vec{v} =$.
The speed is $|\vec{v}| = $.

▶️Answer/Explanation

Solution:

The initial position (at $ t = 0 $) of the body is $(1, 2)$. It is $\sqrt{1^2 + 2^2} = \sqrt{5} = 2.23 \, \text{m}$ far from the origin.
The position at time $ t = 1 \, \text{sec} $ is $(4, 6)$. It is $\sqrt{4^2 + 6^2} = \sqrt{52} = 7.21 \, \text{m}$ far from the origin.
The velocity vector is $\vec{v} = \begin{pmatrix} 3 \\ 4 \end{pmatrix}$.
The speed is $|\vec{v}| = \sqrt{3^2 + 4^2} = 5 \, \text{m/sec}$.

NOTE

If $\vec{r} = \vec{a} + \lambda \vec{b}$ is an equation of a line, the direction vector $\vec{b}$ can be substituted by any multiple of $\vec{b}$.
However, if $\vec{r} = \vec{a} + t\vec{b}$ is an equation of motion, the velocity vector $\vec{b}$ cannot be substituted by a multiple of $\vec{b}$.

Explanation:

Suppose a body is initially at position $ A(1, 2) $ and after 1 second at position $ B(5, 8) $. Then:
Velocity vector: $\vec{v} = \overrightarrow{AB} = \begin{pmatrix} 4 \\ 6 \end{pmatrix}$.
Equation of motion: $\vec{r} = \begin{pmatrix} 1 \\ 2 \end{pmatrix} + t \begin{pmatrix} 4 \\ 6 \end{pmatrix}$.

Suppose the body is initially at $ A(1, 2) $ and after 2 seconds at $ B(5, 8) $. Then:

The direction vector $\vec{b} = \overrightarrow{AB} = \begin{pmatrix} 4 \\ 6 \end{pmatrix}$ corresponds to 2 seconds.
Velocity vector: $\vec{v} = \frac{1}{2} \vec{b} = \begin{pmatrix} 2 \\ 3 \end{pmatrix}$.
Equation of motion: $\vec{r} = \begin{pmatrix} 1 \\ 2 \end{pmatrix} + t \begin{pmatrix} 2 \\ 3 \end{pmatrix}$.

Example

Suppose that a body is moving on a straight line (in 3D space) in the direction of the vector

$\vec{b} = \begin{pmatrix} 1 \\ 2 \\ 2 \end{pmatrix}$ with speed $ 15 \, \text{ms}^{-1} $.

Its initial position is $ A(1, 1, 1) $. Find the equation of the motion of the body.

▶️Answer/Explanation

Solution:

The magnitude of $\vec{b}$ is $|\vec{b}| = \sqrt{1^2 + 2^2 + 2^2} = 3$.
The unit vector is $\hat{b} = \frac{1}{3} \begin{pmatrix} 1 \\ 2 \\ 2 \end{pmatrix}$.
Since the speed is 15, the velocity vector is:
$\vec{v} = 15 \hat{b} = 5 \begin{pmatrix} 1 \\ 2 \\ 2 \end{pmatrix} = \begin{pmatrix} 5 \\ 10 \\ 10 \end{pmatrix}.$
Therefore, the equation of motion is:
$\vec{r} = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} + t \begin{pmatrix} 5 \\ 10 \\ 10 \end{pmatrix}.$

Relative Position Vector $ \vec{AB} $

The vector $ \vec{AB} $ represents the relative position of point B from point A. It tells you how to move from point A to point B using a directed vector.

If $ A(x_1, y_1) $ and $ B(x_2, y_2) $, then:

$ \vec{AB} = \vec{OB} – \vec{OA} = \begin{bmatrix} x_2 – x_1 \\ y_2 – y_1 \end{bmatrix} $

In 3D, if $ A(x_1, y_1, z_1) $, $ B(x_2, y_2, z_2) $, then:

$ \vec{AB} = \begin{bmatrix} x_2 – x_1 \\ y_2 – y_1 \\ z_2 – z_1 \end{bmatrix} $

Key Properties:

Direction: From A to B.
Displacement: Describes the net movement from A to B.
Magnitude: The distance between points A and B:
$ |\vec{AB}| = \sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2} $

Tips:

Always subtract A from B to find $ \vec{AB} $.
The vector $ \vec{BA} $ would just be the negative: $ \vec{BA} = -\vec{AB} $.
Visualize using coordinate axes or triangles if unsure of direction.

Example:

Let $ A(2, -1) $ and $ B(5, 3) $. Find the vector $ \vec{AB} $, which represents the relative position of point B from point A.

▶️ Answer/Explanation

Solution:

The vector $ \vec{AB} $ represents the relative position of point B from point A. It is calculated by subtracting the coordinates of A from B:

$ \vec{AB} = \vec{OB} – \vec{OA} = \begin{bmatrix} 5 \\ 3 \end{bmatrix} – \begin{bmatrix} 2 \\ -1 \end{bmatrix} = \begin{bmatrix} 5 – 2 \\ 3 – (-1) \end{bmatrix} = \begin{bmatrix} 3 \\ 4 \end{bmatrix} $

Therefore, the relative position vector from A to B is:

$ \vec{AB} = \begin{bmatrix} 3 \\ 4 \end{bmatrix} $

Until now, the direction vector $\vec{b}$ (or the velocity vector $\vec{v}$) was a constant vector. For example, if:

$\vec{r} = \begin{pmatrix} 1 \\ 2 \end{pmatrix} + t \begin{pmatrix} 3 \\ 4 \end{pmatrix},$

the velocity vector is $\vec{v} = \begin{pmatrix} 3 \\ 4 \end{pmatrix}$ (constant).

An alternative way to find the velocity vector is by using derivatives. If the motion is given by:

$\vec{r} = \begin{pmatrix} x(t) \\ y(t) \end{pmatrix},$

then the velocity vector is:

$\vec{v} = \frac{d\vec{r}}{dt} = \begin{pmatrix} \dot{x}(t) \\ \dot{y}(t) \end{pmatrix}.$

Similarly, the acceleration is:

$\vec{a} = \frac{d\vec{v}}{dt} = \frac{d^2\vec{r}}{dt^2} = \begin{pmatrix} \ddot{x}(t) \\ \ddot{y}(t) \end{pmatrix}.$

In the case of constant velocity, $\vec{a} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}$.

Example

The equation of motion for a body is given by:

$\vec{r} = \begin{pmatrix} 1 + 3t \\ 2 + 4t + 5t^2 \end{pmatrix}.$

Find

Velocity: $\vec{v} = $.
Acceleration: $\vec{a} = $

At $ t = 0 $:
At $ t = 1 $:

▶️Answer/Explanation

Solution:

Velocity: $\vec{v} = \frac{d\vec{r}}{dt} = \begin{pmatrix} 3 \\ 4 + 10t \end{pmatrix}$.
Acceleration: $\vec{a} = \frac{d\vec{v}}{dt} = \begin{pmatrix} 0 \\ 10 \end{pmatrix}$ (constant).

At $ t = 0 $:
Initial position: $\vec{r} = \begin{pmatrix} 1 \\ 2 \end{pmatrix}$.
Initial velocity: $\vec{v} = \begin{pmatrix} 3 \\ 4 \end{pmatrix}$.
Initial speed: $|\vec{v}| = \sqrt{3^2 + 4^2} = 5 \, \text{ms}^{-1}$.

After 1 second:
Position: $\vec{r} = \begin{pmatrix} 4 \\ 11 \end{pmatrix}$.
Velocity: $\vec{v} = \begin{pmatrix} 3 \\ 14 \end{pmatrix}$.
Acceleration: $\vec{a} = \begin{pmatrix} 0 \\ 10 \end{pmatrix}$.
Speed: $|\vec{v}| = \sqrt{3^2 + 14^2} = \sqrt{205} = 14.3 \, \text{ms}^{-1}$.

Example

If:

$\vec{r} = \begin{pmatrix} 1 + e^{2t} \\ 2 + t^3 \end{pmatrix},$

Find

Velocity:
Acceleration:

▶️Answer/Explanation

Solution:

Velocity: $\vec{v} = \frac{d\vec{r}}{dt} = \begin{pmatrix} 2e^{2t} \\ 3t^2 \end{pmatrix}$.
Acceleration: $\vec{a} = \frac{d\vec{v}}{dt} = \begin{pmatrix} 4e^{2t} \\ 6t \end{pmatrix}$.

◆ Projectile Motion

Projectile motion refers to the motion of an object that is thrown or projected into the air and is subject only to acceleration due to gravity. It is analyzed in two independent components: horizontal and vertical.

Velocity Components:

If a projectile is launched at speed $ v_0 $ and angle $ \theta $ to the horizontal:

\[ v_{x} = v_0 \cos \theta, \quad v_{y} = v_0 \sin \theta \]

Position Equations:

\[ x(t) = v_0 \cos \theta \cdot t \] \[ y(t) = v_0 \sin \theta \cdot t – \frac{1}{2}gt^2 \]

Key Characteristics:

Time of Flight: $ T = \frac{2v_0 \sin \theta}{g} $
Maximum Height: $ H = \frac{(v_0 \sin \theta)^2}{2g} $
Range: $ R = \frac{v_0^2 \sin 2\theta}{g} $

◆ CIRCULAR MOTION

Suppose that:

$\vec{r} = \begin{pmatrix} r \cos \omega t \\ r \sin \omega t \end{pmatrix}.$

Then:

$x^2 + y^2 = (r \cos \omega t)^2 + (r \sin \omega t)^2 = r^2 (\cos^2 \omega t + \sin^2 \omega t) = r^2.$

The equation $ x^2 + y^2 = r^2 $ represents a circle with center $ O(0, 0) $ and radius $ r $.

Thus, the body is moving on a circular path.

Example

If:

$\vec{r} = \begin{pmatrix} 2 \cos 0.1t \\ 2 \sin 0.1t \end{pmatrix},$

Find

Path:
Velocity:
Acceleration:
Speed:

▶️Answer/Explanation

Solution:

Path: $ x^2 + y^2 = 4 $ (circle of radius 2).
Velocity: $\vec{v} = \frac{d\vec{r}}{dt} = \begin{pmatrix} -0.2 \sin 0.1t \\ 0.2 \cos 0.1t \end{pmatrix}$.
Acceleration: $\vec{a} = \frac{d\vec{v}}{dt} = \begin{pmatrix} -0.02 \cos 0.1t \\ -0.02 \sin 0.1t \end{pmatrix}$.
Speed: $|\vec{v}| = \sqrt{(-0.2 \sin 0.1t)^2 + (0.2 \cos 0.1t)^2} = 0.2 \, \text{ms}^{-1}$.

◆Time Shifts and Phase Shifts

In motion equations, a time shift is applied by replacing $ t $ with $ (t – a) $. This shifts the entire graph or motion forward (if $ a > 0 $) or backward (if $ a < 0 $) in time.

General Form:

$ f(t – a) $

Means the original motion $ f(t) $ is shifted a units forward in time.

Application to Circular Motion:

If the position is: $ x(t) = \cos(\omega t) \quad \text{then} \quad x(t – a) = \cos[\omega(t – a)] = \cos(\omega t – \omega a)$

This is also known as a phase shift of $ \omega a $.

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